## 51.20 A bit of uniformity, I

The main task of this section is to formulate and prove Lemma 51.20.2.

Lemma 51.20.1. Let $R$ be a ring. Let $M \to M'$ be a map of $R$-modules with $M$ of finite presentation such that $\text{Tor}_1^ R(M, N) \to \text{Tor}_1^ R(M', N)$ is zero for all $R$-modules $N$. Then $M \to M'$ factors through a free $R$-module.

Proof. We may choose a map of short exact sequences

$\xymatrix{ 0 \ar[r] & K \ar[r] \ar[d] & R^{\oplus r} \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K' \ar[r] & \bigoplus _{i \in I} R \ar[r] & M' \ar[r] & 0 }$

whose right vertical arrow is the given map. We can factor this map through the short exact sequence

51.20.1.1
\begin{equation} \label{local-cohomology-equation-pushout} 0 \to K' \to E \to M \to 0 \end{equation}

which is the pushout of the first short exact sequence by $K \to K'$. By a diagram chase we see that the assumption in the lemma implies that the boundary map $\text{Tor}_1^ R(M, N) \to K' \otimes _ R N$ induced by (51.20.1.1) is zero, i.e., the sequence (51.20.1.1) is universally exact. This implies by Algebra, Lemma 10.82.4 that (51.20.1.1) is split (this is where we use that $M$ is of finite presentation). Hence the map $M \to M'$ factors through $\bigoplus _{i \in I} R$ and we win. $\square$

Lemma 51.20.2. Let $R$ be a ring. Let $\alpha : M \to M'$ be a map of $R$-modules. Let $P_\bullet \to M$ and $P'_\bullet \to M'$ be resolutions by projective $R$-modules. Let $e \geq 0$ be an integer. Consider the following conditions

1. We can find a map of complexes $a_\bullet : P_\bullet \to P'_\bullet$ inducing $\alpha$ on cohomology with $a_ i = 0$ for $i > e$.

2. We can find a map of complexes $a_\bullet : P_\bullet \to P'_\bullet$ inducing $\alpha$ on cohomology with $a_{e + 1} = 0$.

3. The map $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ is zero for all $R$-modules $N$ and $i > e$.

4. The map $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M, N)$ is zero for all $R$-modules $N$.

5. Let $N = \mathop{\mathrm{Im}}(P'_{e + 1} \to P'_ e)$ and denote $\xi \in \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N)$ the canonical element (see proof). Then $\xi$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M, N)$.

6. The map $\text{Tor}_ i^ R(M, N) \to \text{Tor}_ i^ R(M', N)$ is zero for all $R$-modules $N$ and $i > e$.

7. The map $\text{Tor}_{e + 1}^ R(M, N) \to \text{Tor}_{e + 1}^ R(M', N)$ is zero for all $R$-modules $N$.

Then we always have the implications

$(1) \Leftrightarrow (2) \Leftrightarrow (3) \Leftrightarrow (4) \Leftrightarrow (5) \Rightarrow (6) \Leftrightarrow (7)$

If $M$ is $(-e - 1)$-pseudo-coherent (for example if $R$ is Noetherian and $M$ is a finite $R$-module), then all conditions are equivalent.

Proof. It is clear that (2) implies (1). If $a_\bullet$ is as in (1), then we can consider the map of complexes $a'_\bullet : P_\bullet \to P'_\bullet$ with $a'_ i = a_ i$ for $i \leq e + 1$ and $a'_ i = 0$ for $i \geq e + 1$ to get a map of complexes as in (2). Thus (1) is equivalent to (2).

By the construction of the $\mathop{\mathrm{Ext}}\nolimits$ and $\text{Tor}$ functors using resolutions (Algebra, Sections 10.71 and 10.75) we see that (1) and (2) imply all of the other conditions.

It is clear that (3) implies (4) implies (5). Let $N$ be as in (5). The canonical map $\tilde\xi : P'_{e + 1} \to N$ precomposed with $P'_{e + 2} \to P'_{e + 1}$ is zero. Hence we may consider the class $\xi$ of $\tilde\xi$ in

$\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N) = \frac{\mathop{\mathrm{Ker}}(\mathop{\mathrm{Hom}}\nolimits (P'_{e + 1}, N \to \mathop{\mathrm{Hom}}\nolimits (P'_{e + 2}, N)}{ \mathop{\mathrm{Im}}(\mathop{\mathrm{Hom}}\nolimits (P'_ e, N \to \mathop{\mathrm{Hom}}\nolimits (P'_{e + 1}, N)}$

Choose a map of complexes $a_\bullet : P_\bullet \to P'_\bullet$ lifting $\alpha$, see Derived Categories, Lemma 13.19.6. If $\xi$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ R(M', N)$, then we find a map $\varphi : P_ e \to N$ such that $\tilde\xi \circ a_{e + 1} = \varphi \circ d$. Thus we obtain a map of complexes

$\xymatrix{ \ldots \ar[r] & P_{e + 1} \ar[r] \ar[d]^0 & P_ e \ar[r] \ar[d]^{a_ e - \varphi } & P_{e - 1} \ar[r] \ar[d]^{a_{e - 1}} & \ldots \\ \ldots \ar[r] & P'_{e + 1} \ar[r] & P'_ e \ar[r] & P'_{e - 1} \ar[r] & \ldots }$

as in (2). Hence (1) – (5) are equivalent.

The equivalence of (6) and (7) follows from dimension shifting; we omit the details.

Assume $M$ is $(-e - 1)$-pseudo-coherent. (The parenthetical statement in the lemma follows from More on Algebra, Lemma 15.64.17.) We will show that (7) implies (4) which finishes the proof. We will use induction on $e$. The base case is $e = 0$. Then $M$ is of finite presentation by More on Algebra, Lemma 15.64.4 and we conclude from Lemma 51.20.1 that $M \to M'$ factors through a free module. Of course if $M \to M'$ factors through a free module, then $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M', N) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ is zero for all $i > 0$ as desired. Assume $e > 0$. We may choose a map of short exact sequences

$\xymatrix{ 0 \ar[r] & K \ar[r] \ar[d] & R^{\oplus r} \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K' \ar[r] & \bigoplus _{i \in I} R \ar[r] & M' \ar[r] & 0 }$

whose right vertical arrow is the given map. We obtain $\text{Tor}_{i + 1}^ R(M, N) = \text{Tor}^ R_ i(K, N)$ and $\mathop{\mathrm{Ext}}\nolimits ^{i + 1}_ R(M, N) = \mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N)$ for $i \geq 1$ and all $R$-modules $N$ and similarly for $M', K'$. Hence we see that $\text{Tor}_ e^ R(K, N) \to \text{Tor}_ e^ R(K', N)$ is zero for all $R$-modules $N$. By More on Algebra, Lemma 15.64.2 we see that $K$ is $(-e)$-pseudo-coherent. By induction we conclude that $\mathop{\mathrm{Ext}}\nolimits ^ e(K', N) \to \mathop{\mathrm{Ext}}\nolimits ^ e(K, N)$ is zero for all $R$-modules $N$, which gives what we want. $\square$

Lemma 51.20.3. Let $I$ be an ideal of a Noetherian ring $A$. For all $n \geq 1$ there exists an $m > n$ such that the map $A/I^ m \to A/I^ n$ satisfies the equivalent conditions of Lemma 51.20.2 with $e = \text{cd}(A, I)$.

Proof. Let $\xi \in \mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ A(A/I^ n, N)$ be the element constructed in Lemma 51.20.2 part (5). Since $e = \text{cd}(A, I)$ we have $0 = H^{e + 1}_ Z(N) = H^{e + 1}_ I(N) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^{e + 1}(A/I^ m, N)$ by Dualizing Complexes, Lemmas 47.10.1 and 47.8.2. Thus we may pick $m \geq n$ such that $\xi$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^{e + 1}_ A(A/I^ m, N)$ as desired. $\square$

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